Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$

Question

I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$ This solution isn't true. Where am I wrong?

EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.

Answer 1

Integration by parts $$\int x \sqrt{1-x^2} \arcsin x\, dx$$ take $$u'(x)=x \sqrt{1-x^2};\;v(x)=\arcsin x$$ so $$u(x)=\int x \sqrt{1-x^2} \, dx=-\frac{1}{2}\int (-2x)(1-x^2)^{-1/2}\, dx=-\frac{1}{3} \left(1-x^2\right)^{3/2}+C$$ and $$v'(x)=\frac{1}{\sqrt{1-x^2}}$$ apply IBP formula $$\int u'(x)v(x)\,dx=u(x)v(x)-\int u(x)v'(x)\,dx$$

$$\int x \sqrt{1-x^2} \arcsin x\, dx=-\frac{1}{3} \left(1-x^2\right)^{3/2}\arcsin x-\int\left(-\frac{1}{3} \left(1-x^2\right)^{3/2}\right)\frac{1}{\sqrt{1-x^2}}\,dx=$$ $$=-\frac{1}{3} \left(1-x^2\right)^{3/2}\arcsin x+\int \frac{1}{3} \left(1-x^2\right)\,dx=$$ $$=-\frac{1}{3} \left(1-x^2\right)^{3/2} \arcsin x+\frac{1}{3} \left(x-\frac{x^3}{3}\right)+C$$ which can be simplified to $$=\frac{1}{9} \left(3x-x^3-3 \left(1-x^2\right)^{3/2} \arcsin x\right)+C$$

Answer 2

Let $x=\sin(\theta)$.* Then, $$ \frac{dx}{d\theta}=\cos(\theta) \implies dx=\cos(\theta)d\theta \, . $$ The integral becomes $$ \int \sin(\theta) \left(\sqrt{1-\sin(\theta)^2}\right) \theta\cos(\theta)d\theta $$

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*To be more precise, we are making the substitution $\theta=\arcsin(x)$, so that $-\pi/2 \leq \theta \leq \pi/2$. This allows us to conclude that $\arcsin(\sin(\theta))=\sin(\theta)$.

Answer 3

Let $x=\sin(u)$, then $dx=\cos(u)du$. So $$\begin{align}\int x\sqrt{1-x^2}\arcsin(x)dx&=\int u\sin(u)\cos^2(u)du\\ &=\int u\sin(u)du-\int u\sin^3(u)du\end{align}$$

Answer 4

Your mistake is a sign error:$$\begin{align}\frac{d}{dx}x^3(1-x^2)^{3/2}&=x^2(1-x^2)^{1/2}[3(1-x^2)+x\tfrac32(-2x)]\\&=x^2(1-x^2)^{1/2}[3-3x^2-3x^2].\end{align}$$You seem to have mistaken that last $-$ sign for a $+$. Everything after that is error-carried-forward.

Answer 5

$$\int x\sqrt{1-x^2}sin^{-1}x dx$$ $$x = sin\theta$$ $$cos\theta= \frac{dx}{d\theta}$$ $$dx= cos\theta d\theta$$ $$\int x\sqrt{1-x^2}sin^{-1}x dx= \int sin\theta(\sqrt{cos^2\theta})\theta (cos\theta d\theta)$$ $$=\int \theta sin\theta cos^2\theta d\theta$$ $$=\int \theta sin\theta (1-sin^2\theta) d\theta$$ $$=\int \theta sin\theta -\int \theta sin^3\theta d\theta$$ $$=\int \theta sin\theta -\int \theta\Bigl(\frac{3sin\theta}{4}-\frac{sin3\theta}{4}\Bigl) d\theta$$ $$=\int \theta sin\theta -\int\frac{3\theta sin\theta}{4}+ \int\frac{\theta sin3\theta}{4}d\theta$$ Applying product rule, $$\int \theta sin\theta= \theta \int sin\theta - \int(\theta)' \int sin\theta d\theta= sin\theta - \theta cos\theta$$ $$-\int\frac{3\theta sin\theta}{4}= \frac{3\theta cos\theta}{4} -\frac{3 \sin\theta}{4}$$ $$\int\frac{\theta sin3\theta}{4}d\theta=\frac{1}{4}[\frac{sin3\theta}{9}-\frac{\theta cos3\theta}{3}]$$ $$\int x\sqrt{1-x^2}sin^{-1}x dx=sin\theta - \theta cos\theta\ + \frac{3\theta cos\theta}{4} -\frac{3 \sin\theta}{4}+\frac{sin3\theta}{36}-\frac{\theta cos3\theta}{12}$$ $$= \frac{1}{4}(sin \theta - \theta cos\theta)+ \frac{sin3\theta}{36}-\frac{\theta cos3\theta}{12}+ c$$ Substitute theta with corresponding expression in x

Answer 6

Let $x=\sin{u}$. Substitute $\arcsin{x}=u$ to change variables, notice that $d(\cos^3{u})=-3\sin{u}\cos^2{u}\,du$ and get the integral by parts $$ \int x\sqrt{1-x^2}\arcsin{x}\,dx = \int{u}\,(\sin{u}\cos^2{u}\,du) = -\frac{1}{3}\int{u}\,d(\cos^3{u}) = \frac{1}{3}\Bigl(\int\cos^3{u}\,du - u\cos^3{u} \Bigr) $$ After back substitution the last integral looks $$ \frac{1}{3}\Bigl(\int(1-x^2)\,dx - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr) = \frac{1}{3}\Bigl( x - \frac{x^3}{3} - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr) $$ Don't forget a constant and get the answer $$ \int{x\sqrt{1-x^2}\arcsin{x}}\,dx = \frac{1}{3}\Bigl( x - \frac{x^3}{3} - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr) + C $$ Check the math and good luck!